transgalactic
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prove that..
[tex] x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\[/tex]
[tex] R_5=\frac{f^{5}(c)x^5}{5!}[/tex]
i need to prove that the remainder is negative .
[tex] \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5[/tex]
[tex] R_5=\frac{cos(c)x^5}{5!}[/tex]
[tex] x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\[/tex]
[tex] R_5=\frac{f^{5}(c)x^5}{5!}[/tex]
i need to prove that the remainder is negative .
[tex] \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5[/tex]
[tex] R_5=\frac{cos(c)x^5}{5!}[/tex]